Formal Logic 101 – Part 7: Rules of Inference (Part 2)

Last time we re-visited two inference rules and introduced three more. This time around we’ll cover three more inference rules, and then we’ll just have two more. First on the list we have:

Disjunctive Syllogism (DS)

This is our first inference rule dealing with “Or” as an operator. Disjunctive Syllogism states: Given the premises “AvB” and “~A”, conclude “B”
1. AvB            Premise (P)
2. ~A              P
3. B                 1,2 DS

Recall that an Or statement states “At least one of the two propositions is true.” So if we know that either “A” or “B” must be true, and we know it’s not “A”, then simple process of elimination leaves us with only one possible answer: B.

Addition (Add)

Addition is something of the opposite of a Disjunctive Syllogism. I will say up front that – while this inference rule is not difficult – it is one that people tend to instinctively dislike. Personally, I think it’s because the rule feels like you’re intentionally muddying the waters, but it IS valid, and for some proofs it IS useful for ensuring that there’s a clear series of steps leading from the premises to the conclusion.

Addition states: Given the premise “A”, conclude “AvB”.
1. A              Premise
2. AvB         1 Add.

As an example consider the phrase “Tom is an axe-murderer”. "Addition" states that we can take this premise and say, “Either Tom is an axe-murderer, or Sarah is”. This is usually the part where people start instinctively rebelling against this rule. After all, we’ve already established that Tom is the axe-murderer. Why are we dragging Sarah into this? Why are we seemingly implying that it might not be Tom when we’ve already said, “It’s Tom”? Once again, I will ask you to remember that “Or” statements tell us “At least one of the two propositions is true” and our premise (which we always assume to be true) is one of the propositions in our “Or” statement, thus meeting the requirement that at least one proposition of the “Or” statement be true.

Using addition and one proposition, we can conclude virtually any “Or” statement imaginable. It is a perfectly valid logical step to move from “this Crayon is red” to “either this Crayon is red or the Moon Landing was a hoax”. Because Addition can be used so broadly, it’s generally a bad idea to use it willy-nilly (unless you just really really want to have a 500-line proof where you don’t use 480 of the lines for anything). It’s generally recommended that you have a goal in mind and know what you hope to accomplish by using Addition before using the rule.

Dilemma (Dil)

Our last basic inference rule will be Dilemma. This will also be our only basic rule that requires 3 premises to justify it. Dilemma states: Given the premises “AvB”, “A→C”, and “B→D”, conclude “CvD”.

1. AvB         Premise (P)
2. A→C       P
3. B→D       P
4. CvD         1,2,3 Dil.

As a concrete example consider the recent presidential election. Let’s say we’ve been living under a rock. We know that either Biden or Trump won the election, but we don’t know who. We know if Biden won, Harris is the Vice President. We know if Trump won, Pence is the Vice President. So we can conclude that either Harris or Pence is the Vice President.

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Restrictions on Use

To wrap up this post, I want to talk about a restriction that will apply to every rule we have discussed thus far: You can only use these rules on whole lines of a proof. What does this mean? Let me draw an example:

If my Identity was stolen, I will need a new Social security card and a new credit card.
This is a three-proposition sentence. Let “I” be “my Identity was stolen”, “S” be “I need a new Social security card”, and “C” be “I need a new Credit card”. So the sentence is correctly symbolized as.
1. I→(S^C)        Premise (P).

I cannot from this draw the conclusion:
2. C            1 Simp. INVALID

This is because premise 1 does not state “I need a new social security and credit card”; it only states that I WOULD need them if my identity were stolen. We’re still hanging out in purely hypothetical territory, and we cannot conclude a fact such as “I need a new credit card” from hypotheticals.

Now consider what changes if we introduce one more premise:
1. I→(S^C)        Premise (P).
2. I                      P – (prove “C”)
From here, we can use Modus Ponens to conclude:
3. S^C               1,2, MP
And from THERE we can use simplification on line 3 to reach our conclusion:
4. C                  3 simp.

As a rule, if you see an operator that is inside of parentheses, assume that you cannot do anything with it (at least not with any inference rules) until you get it out of the parentheses.

So that’s it. Those are our basic inference rules. Next time we’ll learn one of our two slightly more complex inference rules.



Practice

Using all of the rules we’ve learned so far, complete the following proofs. Remember to list the rules you use, so that you can double-check yourself later.



Argument A
1. PvD                                    Premise (P)
2. (P→ ~A) ^ (E→ ~D)         P
3. E                                         P – Prove ~A

Note: If you find yourself overwhelmed, try finding some phrases to help you translate the phrases back to regular English. You can generally pick any phrases you like. For instance, I might say something like:

1. That creature is either a Puppy or a Dolphin.
2. If it’s a Puppy then it’s not Aquatic, and if it has large Ears then it’s not a Dolphin.
3. That creature has large Ears.

And then attempt to prove that the creature is not Aquatic.

Argument B
1. [(SvD)→~W]^[(~Wv~E)→R]        P
2. R→(~X^~T)                                    P
3. S                                                      P – Prove ~T

Argument C
1. P→[Q→(RvS)         P
2. P^Q                          P
3. S→T                        P
4. ~Tv~W                    P
5. ~~W                         P – Prove R

This last one will be a bit more challenging.

Argument D
1. (~A^~B)→(~Cv~D)      P
2. (Ev~F)→~A                   P
3. ~H→(B→J)                   P
4. (~F^~H)→(~~C^~J)    P
5. ~H^(F→H)                     P – Prove ~D

Answers

Note that solutions provided are not necessarily the only possible solutions.

Argument A
1. PvD                              Premise (P)
2. (P→~A) ^ (E→~D)     P
3. E                                   P – Prove ~A
4. E→~D                          2 simp
5. ~D                                3,4 MP
6. P                                   1,5 DS
7. P→~A                          2 simp
8. ~A                                6,7 MP

Argument B:
1. [(SvD)→~W]^[(~Wv~E)→R]        Premise (P)
2. R→(~X^~T)                                    P
3. S                                                       P – Prove ~T
4. (SvD)→~W                                     1 simp.
5. SvD                                                   3 add
6. ~W                                                   4,5 MP
7. (~Wv~E)→R                                  1 simp.
8. (~Wv~E)→(~X^~T)                       2,7 HS
9. ~Wv~E                                             6 add
10. ~X^~T                                           8,9 MT
11. ~T                                                  10 simp

Argument C
1. P→[Q→(RvS)]           P
2. P^Q                             P
3. S→T                            P
4. ~Tv~W                        P
5. ~~W                            P – Prove R
6. ~T                                4,5 DS
7. ~S                                3,6 MT
8. P                                  2 simp
9. Q                                 2 simp
10. Q→(RvS)                 1,8 MP
11. RvS                           9,10 MP
12. R                               7,11 DS

Argument D
1. (~A^~B)→(~Cv~D)         P
2. (Ev~F)→~A                      P
3. ~H→(B→J)                      P
4. (~F^~H)→(~~C^~J)       P
5. ~H^(F→H)                        P – Prove ~D
6. ~H                                     5 simp
7. F→H                                 5 simp
8. ~F                                      6,7 MT
9. ~F^~H                               6,8 conj
10. ~~C^~J                          4,9 MP
11. Ev~F                               8 add
12. ~A                                   2,11 MP
13. B→J                               3,6 MP
14. ~J                                   10 simp
15. ~B                                  13,14 MT
16. ~A^~B                           12,15 conj
17. ~Cv~D                           1,16, MP
18. ~~C                               10 simp
19. ~D                                  17,18 DS


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